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\usepackage{amsmath, amsthm, amssymb, bm} % 数学公式与符号
\usepackage{graphicx}
\usepackage{hyperref}
\usepackage{booktabs} % 用于高质量表格
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\title{自由落体运动与有空气阻力的落体运动}
\author{五六七}
%\date{2025年9月3日}

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\begin{document}

% 封面页
\begin{frame}
  \titlepage
\end{frame}

% 目录页
\begin{frame}{目录}
  \tableofcontents
\end{frame}

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\section{自由落体运动}
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\begin{frame}[allowframebreaks]{自由落体运动}

问题：自由落体在 $t$ 时刻的高度记为 $y(t)$, 写出 $y(t)$ 满足的微分方程。


\begin{figure}[h]
    \centering
    \begin{tikzpicture}[scale=0.7, >=Stealth, thick]

        % 定义点的位置
        \coordinate (O) at (0,0);

        \coordinate (A) at (0,3);
        \coordinate (B) at (0,4);
        \coordinate (C) at (-1,3);
        \coordinate (D) at (-1,1.5);

        % 绘制地面
        \draw[thick] (-3,0) -- (3,0) node[above] {地面};

        % 绘制向量y轴
        \draw[->, -{Stealth[scale=1.5]}, thick] (O) -- (B) node[above right] {$y$};
        \fill[blue] (A) circle (1.5pt) node[right] {$y=y(t)$};
        \fill[blue] (O) circle (1.5pt) node[above right] {$O$};

        % 绘制直线虚线
        \draw[dashed] (C) -- (A);

        % 绘制物体
        \fill[blue] (C) circle (2.5pt);

        %物理受力分析
        \draw[->, -{Stealth[scale=1.5]}, thick, purple] (C) -- (D) node[below] {$mg$};

    \end{tikzpicture}
    \caption{自由落体运动}
    \label{fig:free-fall}
\end{figure}


解答：

牛顿第二运动定律是 $F=ma$, 其中 $F$ 是合力，$m$ 是质量，$a$ 是加速度。

代入本例，可得 $my{\,}''(t) = -mg$. 化简可得 $y{\,}''(t)=-g$.

积分一次，可得 $y{\,}'(t)=-gt+C_1$, 其中 $C_1$ 是任意常数。

再积分一次，可得 $y(t)=- \frac{1}{2}gt^2+C_1t+C_2$, 其中 $C_2$ 是任意常数。

\newpage

初始高度为 $y(0)=C_2$. 

初始速度为 $y{\,}'(0)=C_1$. 

名词解释：{\color{red}初值问题}，又称{\color{red}柯西问题}，是微分方程+初值条件：
\[
\begin{cases}
    y{\,}''(t) = -g, \\
    y(0) = y_0, y{\,}'(0)=v_0.
\end{cases}
\]


%示例图（请替换实际图像路径）:
% \begin{center}
% \includegraphics[height=0.8\textheight, width=0.9\textwidth]{path/to/your/image.png}
% \end{center}

问题：验证通解 $y(t)=- \frac{1}{2}gt^2+C_1t+C_2$ 中的常数 $C_1,C_2$ 是{\color{red}独立常数}。


\end{frame}

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\section{有空气阻力的落体运动}
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\begin{frame}[allowframebreaks]{物体在空气中的降落}

问题：物体在空气中下落，例如高空跳伞。假设空气阻力与速度的平方成正比。
%设垂直向下的方向为正方向，
在 $t$ 时刻的高度记为 $y(t)$, 写出 $y(t)$ 满足的微分方程。

\begin{figure}[h]
    \centering
    \begin{tikzpicture}[scale=0.6, >=Stealth, thick]

        % 定义点的位置
        \coordinate (O) at (0,0);

        \coordinate (A) at (0,3);
        \coordinate (B) at (0,4);
        \coordinate (C) at (-1,3);
        \coordinate (D) at (-1,1.5);
        \coordinate (E) at (-1,4);

        % 绘制地面
        \draw[thick] (-3,0) -- (3,0) node[above] {地面};

        % 绘制向量y轴
        \draw[->, -{Stealth[scale=1.5]}, thick] (O) -- (B) node[above right] {$y$};
        \fill[blue] (A) circle (1.5pt) node[right] {$y=y(t)$};
        \fill[blue] (O) circle (1.5pt) node[above right] {$O$};

        % 绘制直线虚线
        \draw[dashed] (C) -- (A);

        % 绘制物体
        \fill[blue] (C) circle (2.5pt);

        %物理受力分析
        \draw[->, -{Stealth[scale=1.5]}, thick, purple] (C) -- (D) node[below] {$mg$};
        \draw[->, -{Stealth[scale=1.5]}, thick, purple] (C) -- (E) node[above] {$f$};

    \end{tikzpicture}
    \caption{有空气阻力的落体运动}
    \label{fig:fall-with-air}
\end{figure}

解答：根据牛顿第二运动定律，可得微分方程 $my{\,}''(t) = -mg + ky{\,}'(t)^2$. 

因为方程中未出现 $y(t)$, 记速度 $v(t)=y{\,}'(t)$, 则方程写为 $ mv{\,}'(t) = -mg + kv(t)^2$.

分离变量，可得
$$ \frac{mdv}{-mg + kv^2} = dt. $$

两边积分，可得
$$ \int_{v=v_0}^{v(t)} \frac{dv}{-mg + kv^2} = \int_{t=t_0}^{t} dt. $$

%由此速度 $v(t)$ 和位置 $y(t)$ 的表达式。


微分方程的标准形式
$$ \frac{dv}{dt} = -g + \frac{k}{m}v^2. $$

问题：在 $(t,v)$ 平面画出这个微分方程的{\color{red}线素场}。

%示意图（请替换实际图像路径）:
% \begin{center}
% \includegraphics[height=0.8\textheight, width=0.4\textwidth]{path/to/your/sky-dive-image.png}
% \end{center}

问题：求解这个微分方程，画出{\color{red}积分曲线族}。

%积分曲线族（请替换实际图像路径）:
% \begin{center}
% \includegraphics[height=0.8\textheight, width=0.8\textwidth]{path/to/your/integral-curves-image.png}
% \end{center}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{有空气阻力的落体运动的微分方程的求解}
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\begin{frame}[allowframebreaks]{微分方程求解}

通过变量代换
\[
\begin{cases}
s = -\sqrt{\frac{kg}{m}}t, \\
z = \sqrt{\frac{k}{mg}}v,
\end{cases}
\]
微分方程 
$$ \frac{dv}{dt} = -g + \frac{k}{m}v{\,}^2 $$ 
可化为
$$\frac{dz}{ds} = 1-z{\,}^2.$$

分离变量可得
$$\frac{dz}{1-z{\,}^2} = ds.$$

两边积分，当$|z|<1$ 时（即 $|v|<\sqrt{\frac{mg}{k}}$时），可得
$$\frac{1+z}{1-z} = Ce^{2s},$$
其中$C$ 是任意常数。

由此解得 
$$z= \frac{Ce^{2s}-1}{Ce^{2s}+1}.$$

代回原变量，可得微分方程的解为
$$
\sqrt{\frac{k}{mg}}v = \frac{Ce^{-2\sqrt{\frac{kg}{m}}t}-1}{Ce^{-2\sqrt{\frac{kg}{m}}t}+1}.
$$



\end{frame}

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\end{document}

